Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> dp2[18][100007];
int dp1[18][100007];
vector<int> adj[100007];
vector<int> person[100007];
int depth[100007];
int jump(int u, int diff, vector<int>& ans) {
  while (diff > 0) {
    vector<int> temp;
    temp.resize(20);
    merge(ans.begin(), ans.end(), dp2[(int)log2(diff)][u].begin(),
          dp2[(int)log2(diff)][u].end(), temp.begin());
    temp.resize(min(10, (int)(ans.size() + dp2[(int)log2(diff)][u].size())));
    ans = temp;
    u = dp1[(int)log2(diff)][u];
    diff = diff - (1 << ((int)log2(diff)));
  }
  return u;
}
int lca(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  int diff = depth[u] - depth[v];
  for (int i = 0; i < 18; i++)
    if ((diff >> i) & 1) u = dp1[i][u];
  if (u == v) return u;
  for (int i = 17; i >= 0; i--)
    if (dp1[i][u] != dp1[i][v]) {
      u = dp1[i][u];
      v = dp1[i][v];
    }
  return dp1[0][u];
}
void dfs(int i, int parent, int temp) {
  dp1[0][i] = parent;
  depth[i] = temp;
  for (int j = 0; j < min(10, (int)person[parent].size()); j++)
    dp2[0][i].push_back(person[parent][j]);
  for (int j = 0; j < adj[i].size(); j++)
    if (adj[i][j] != parent) dfs(adj[i][j], i, temp + 1);
}
void gen(int n, int m) {
  dfs(1, -1, 0);
  for (int i = 1; i < 18; i++)
    for (int j = 1; j <= n; j++) {
      if (dp1[i - 1][j] == -1) continue;
      dp1[i][j] = dp1[i - 1][dp1[i - 1][j]];
      dp2[i][j].resize(20);
      merge(dp2[i - 1][j].begin(), dp2[i - 1][j].end(),
            dp2[i - 1][dp1[i - 1][j]].begin(), dp2[i - 1][dp1[i - 1][j]].end(),
            dp2[i][j].begin());
      dp2[i][j].resize(min(
          10, (int)(dp2[i - 1][j].size() + dp2[i - 1][dp1[i - 1][j]].size())));
    }
}
void func(int u, int v, int a, int m) {
  if (depth[u] > depth[v]) swap(u, v);
  vector<int> ans;
  set<int> temp;
  int LCA = lca(u, v);
  int diff1 = depth[u] - depth[LCA];
  int diff2 = depth[v] - depth[LCA] - 1;
  jump(u, diff1, ans);
  if (v != LCA) jump(v, diff2, ans);
  for (int i = 0; i < min(10, (int)person[u].size()); i++)
    temp.insert(person[u][i]);
  for (int i = 0; i < ans.size(); i++) temp.insert(ans[i]);
  if (LCA != v)
    for (int i = 0; i < min(10, (int)person[v].size()); i++)
      temp.insert(person[v][i]);
  a = min((int)temp.size(), a);
  printf("%d ", a);
  set<int>::iterator k = temp.begin();
  for (int i = 0; i < a; i++, k++) printf("%d ", *k);
  printf("\n");
}
int scan() {
  int a;
  scanf("%d", &a);
  return a;
}
int main() {
  int n, m, q;
  n = scan();
  m = scan();
  q = scan();
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    u = scan();
    v = scan();
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  int c[m + 1];
  for (int i = 1; i <= m; i++) {
    c[i] = scan();
    person[c[i]].push_back(i);
  }
  gen(n, m);
  while (q--) {
    int u, v, a;
    u = scan();
    v = scan();
    a = scan();
    func(u, v, a, m);
  }
  return 0;
}